Problem on exponential distribution and conditional probability There are two batteries in the box. The operating time of each of them is exponential distribution with parameters $\lambda_i$, $i=1,2$, The i-th battery is got with probability $p_i$, $p_1+p_2=1$.
It is known that it worked t hours. What is the probability that it will work for another s hours?
I got confused in this problem because I even have solution that seems logical to me, but it gives strange answer and I cannot understand where the error is.
Here are my thoughts:
I) $$F(X<x)=1-\exp(-\lambda_ix)  => F(X>x)=\exp(-\lambda_ix)$$
$$P(t+s>x|t>x)=\frac{P(t>x|t+s>x)P(t+s>x)}{P(t>x)}=\frac{1\exp(-\lambda_i(t+s))}{\exp(-\lambda_i(t)}=\exp(-\lambda_is)$$
$$P=p_1\exp(-\lambda_1s)+p_2\exp(-\lambda_2s)$$
It seems to me very strange that  the final result does not depend on t. It is possible that this is directly related to the exponential distribution property (no memory), but I'm not sure.
I will be glad to any feedback
 A: Let $B_1$ (similarly $B_2$) be the event that battery $1$ (similarly battery $2$) was the one picked. Let $X$ be the lifetime. Then,
\begin{eqnarray*}
P\left(X > t+s\,|\,X>t,B_1\right) &=& e^{-\lambda_1 s},\\
P\left(X > t+s\,|\,X>t,B_2\right) &=& e^{-\lambda_2 s}.
\end{eqnarray*}
We then calculate
\begin{equation*}
P\left(X >t+s\,|\,X>t\right) = P\left(X >t+s\,|\,X>t,B_1\right)P\left(B_1\,|\,X>t\right) + P\left(X >t+s\,|\,X>t,B_2\right)P\left(B_2\,|\,X>t\right).
\end{equation*}
Some of these expressions we have computed. It remains to calculate
\begin{equation*}
P\left(B_1\,|\,X>t\right) = \frac{P\left(X>t\,|\,B_1\right)P\left(B_1\right)}{P\left(X>t\,|\,B_1\right)P\left(B_1\right) + P\left(X>t\,|\,B_2\right)P\left(B_2\right)}.
\end{equation*}
The other conditional probability is found similarly. So, the answer does depend on $s$ and $t$. You use memorylessness to compute probabilities of the form $P\left(X>t+s\,|\,X>t,B_i\right)$, but you still need probabilities of the form $P\left(B_i\,|\,X>t\right)$, which do depend on $t$.
